Defining a subset of uniform convergence of a succession of functions

I am new to convergence analysis of successions of functions, but I'm quite stuck on this particular exercise. $$f_n(x)=-\arctan\left(\frac{nx}{n+x}\right)$$ is a succession of functions.
Study pointwise and uniform convergence, and determine an interval of uniform convergence.

Pointwise Convergence:
$$\lim_{n\to \infty}-arctan(\frac{nx}{n+x})=\lim_{n\to ∞}-\arctan(\frac{x}{1+x/n})=-\arctan(x)$$.
Uniform Convergence:
$$\sup_{x\in\Bbb R}{|f_n(x)-f(x)|}=$$ ?
$$|-arctan(\frac{nx}{n+x})+arctan(x)|\le |-arctan(\frac{nx}{n+x})| + |arctan(x)| \le \pi$$
Therefore:
$$\sup_{x\in\Bbb R}{|f_n(x)-f(x)|}=\pi \Rightarrow$$ No uniform convergence in $$\Bbb R$$.

Looking for a subset of $$\Bbb R$$ where $$f_n(x)$$ converges uniformly: $$\partial_{x}(f_n(x)-f(x))=\frac{x(2n+x)}{(1+x^2)(n^2x^2+(n+x)^2)}$$

As denominator is positive, $$f_n(x)-f(x)$$ is monotone decreasing for $$x \in (-2n,0)$$.
My questions are: how can I now define a subset of $$\Bbb R$$ in which $$f_n(x)$$ converges uniformly? Which point of view (or rule) should I use when facing a similar situation?

You can take, for instance, the interval $$[0,1]$$. In that interval, $$f_n-f$$ is a decreasing function which takes the value $$0$$ at $$0$$. So, the minimum is attained at $$1$$. In other words,$$(\forall x\in[0,1]):0\geqslant f_n(x)-f(x)\geqslant f_n(1)-f(1).$$So, since $$\lim_{n\to\infty}f_n(1)-f(1)=0$$, the convergence is uniform in $$[0,1]$$.
Of course, by the same argument, any interval $$[0,1]$$ (with $$a>0$$) will do.
• Thank you, you have brought some light in my reasoning. In my case, it is correct (due to $f_n - f$ decreasing in $x \in (-2n,0)$ ) to say that there is uniform convergence in $[\varphi,0)$ with $\varphi \lt 0, \forall \varphi \in \Bbb R$ as $|f_n(\varphi)-f(\varphi)|_{n \rightarrow \infty} \rightarrow 0$ and $|f_n(0)-f(0)|_{n \rightarrow \infty} \rightarrow 0$ ? Commented May 10, 2019 at 19:29
• There's a subtlety there. Note that $f_n$ is undefined at $-n$. However, if $N$ is large enough, then the sequence $(f_n)_{n\geqslant N}$ converges uniformly ion $[\varphi,0]$. Commented May 10, 2019 at 20:05